Question

The Python datetime.isocalendar() method returns a tuple (ISO_year, ISO_week_number, ISO_weekday) for the given datetime object. Is there a corresponding inverse function? If not, is there an easy way to compute a date given a year, week number and day of the week?

Answer 1

Python 3.8 added the fromisocalendar() method:

>>> datetime.fromisocalendar(2011, 22, 1)
datetime.datetime(2011, 5, 30, 0, 0)

Python 3.6 added the %G, %V and %u directives:

>>> datetime.strptime('2011 22 1', '%G %V %u')
datetime.datetime(2011, 5, 30, 0, 0)

Original answer

I recently had to solve this problem myself, and came up with this solution:

import datetime

def iso_year_start(iso_year):
    "The gregorian calendar date of the first day of the given ISO year"
    fourth_jan = datetime.date(iso_year, 1, 4)
    delta = datetime.timedelta(fourth_jan.isoweekday()-1)
    return fourth_jan - delta 

def iso_to_gregorian(iso_year, iso_week, iso_day):
    "Gregorian calendar date for the given ISO year, week and day"
    year_start = iso_year_start(iso_year)
    return year_start + datetime.timedelta(days=iso_day-1, weeks=iso_week-1)

A few test cases:

>>> iso = datetime.date(2005, 1, 1).isocalendar()
>>> iso
(2004, 53, 6)
>>> iso_to_gregorian(*iso)
datetime.date(2005, 1, 1)

>>> iso = datetime.date(2010, 1, 4).isocalendar()    
>>> iso
(2010, 1, 1)
>>> iso_to_gregorian(*iso)
datetime.date(2010, 1, 4)

>>> iso = datetime.date(2010, 1, 3).isocalendar()
>>> iso
(2009, 53, 7)
>>> iso_to_gregorian(*iso)
datetime.date(2010, 1, 3)